Chapter 3 x Integral Relations for a Control Volume
191
velocity, decaying vertically above the wake according to the variation u | Uo 'Uez/L, where
L is the plate height and z 0 is the top of the wake. Find 'U as a function of stream speed Uo.
Fig. P3.25
Solution: For a control volume enclosing the upper half of the plate and the section
where the exponential profile applies, extending upward to a large distance H such that
exp(–H/L) | 0, we must have inlet and outlet volume flows the same:
H
Q in
³
H
U o dz
L/2
L·
§
) dz, or: U o ¨ H ¸
2¹
©
0
1
Cancel U o H and solve for 'U | U o Ans.
2
³ (Uo 'U e
Q out
z/L
U o H 'UL
3.26 A thin layer of liquid, draining from
an inclined plane, as in the figure, will have
a laminar velocity profile u Uo(2y/h y2/h2),
where Uo is the surface velocity. If the
plane has width b into the paper, (a) determine the volume rate of flow of the film.
(b) Suppose that h 0.5 in and the flow
rate per foot of channel width is 1.25 gal/min.
Estimate Uo in ft/s.
Solution: (a) The total volume flow is computed by integration over the flow area:
Q
³ Vn dA
§ 2 y y2 ·
U
³ o ¨© h h2 ¸¹ b dy
0
h
2
U o bh
3
Ans. (a)
(b) Evaluate the above expression for the given data:
gal
Q 1.25
min
ft 3
0.002785
s
solve for U o
2
2
§ 0.5 ·
Uo bh
U o (1.0 ft ) ¨
ft ¸ ,
3
3
© 12 ¹
ft
0.10
Ans. (b)
s